Question

Solve the cube root equation

\sqrt[3]{x+28} - \sqrt[3]{x-28} = 2

Collected in the board: Cube root

Steven Zheng posted 5 days ago

Answer

Let

a = \sqrt[3]{x+28}
(1)
b = \sqrt[3]{x-28}
(2)

Then

a - b = 2
(3)

Difference of a and b raised to the power of 3 gives

a^3 - b^3 = 56
(4)

Factor LHS using difference of cubes formula

(a - b)(a^2+ab+b^2) = 56
(5)

Substitute (3) to (5) and simplify

a^2+ab+b^2 = 28
(6)

Square of equation (3) gives

(a -b )^2 = 4

Expand LHS

a^2 - 2ab +b^2 = 4
(7)

Substract (7) from (6) and solve for ab

ab = \dfrac{1}{3}(\dfrac{2\cdot 28}{2} - 2^2 ) = 8
(8)

Substituting ab to (6) gives

a^2+b^2 = 28 - ab = 20
(9)

Construct a perfect square binomial with (9) and (8)

(a+b)^2 = a^2+b^2+2ab
(a+b)^2 = 20 +2\cdot 8 = 36

Taking square root of both sides gives

a +b = \pm6
(10)

Now we get two systems of equations

Case 1

\begin{cases} a+b = 6 \\ a - b =2 \end{cases}

which gives

\begin{cases} a= 4 \\ b =5 \end{cases}

Then

x = a^3 - 28 = 4^3 - 28 = 36

Case 2

\begin{cases} a+b = -6 \\ a - b =2 \end{cases}

which gives

\begin{cases} a= -2 \\ b =-4 \end{cases}

Then

x = a^3 - 28 = -2^3 - 28 = -36

In summary, we get two solutions for x.

x = 36 \text{ or }x = -36


Steven Zheng posted 5 days ago

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